According to the second law you can only get a system to do "work" if parts of the system are at different temperatures. In this situation the system is a diode and it does work by converting heat into light. It is hard to tell from the description, but I am guessing the entire diode is at an elevated temperature.
harry On Tue, Feb 28, 2012 at 12:20 PM, Daniel Rocha <danieldi...@gmail.com> wrote: > Why do you think it would violate the 2nd law? I don't understand. > > 2012/2/28 Harry Veeder <hveeder...@gmail.com> >> >> On Tue, Feb 28, 2012 at 11:57 AM, Daniel Rocha <danieldi...@gmail.com> >> wrote: >> > Pay attention at this: >> > >> > " Experiments directly confirm for the first time that this behavior >> > continues beyond the conventional limit of unity electrical-to-optical >> > power >> > conversion efficiency." >> > >> > It is above the conventional, not that it produces energy out of >> > nothing. >> > This is just a way of saying that it exceeded expectation of light >> > emission >> > for a LED. >> >> >> Yes. It uses electricity to change heat into light. The abstract: >> >> "A heated semiconductor light-emitting diode at low forward bias >> voltage V<kBT/q is shown to use electrical work to pump heat from the >> lattice to the photon field. Here the rates of both radiative and >> nonradiative recombination have contributions at linear order in V. As >> a result the device’s wall-plug (i.e., power conversion) efficiency is >> inversely proportional to its output power and diverges as V >> approaches zero. Experiments directly confirm for the first time that >> this behavior continues beyond the conventional limit of unity >> electrical-to-optical power conversion efficiency." >> >> >> however, wouldn't this require a violation of the second law of >> thermodynamics? >> >> Harry >> > > > > -- > Daniel Rocha - RJ > danieldi...@gmail.com >
